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  7. Introduction to Computer Engineering (4023CEM)

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Subnettin
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  • Course
  • Introduction to Computer Engineering (4023CEM)
  • Institution
  • Coventry University (West Midlands) (CU)

Suppose that a router interconnects three subnets, where all the interfaces of the subnets must have the prefix 164.132.62/24. Also, suppose that Subnet 1 is required to support up to 60 interfaces, Subnet 2 is required to support up to 100 interfaces, and Subnet 3 is required to support up to 15 i...

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Document information

  • May 23, 2021
  • 3
  • 2020/2021
  • Exam (elaborations)
  • Questions & answers

Subjects

  • prefix
  • subnet
  • network addresses
  • explain and apply the basics of computer networks and tcpip protocols
  • router interconnects three subnets
  • abcdx

Written for

  • Coventry University (West Midlands) (CU)
  • Coventry University
  • Introduction to Computer Engineering (4023CEM)
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Content preview

Suppose that a router interconnects three subnets, where all the interfaces of
the subnets must have the prefix 164.132.62/24. Also, suppose that Subnet 1 is
required to support up to 60 interfaces, Subnet 2 is required to support up to
100 interfaces, and Subnet 3 is required to support up to 15 interfaces. Provide
three network addresses of the form a.b.c.d/x that satisfy the above
requirements.



We have three subnets that must have the prefix 164.132.62/24. Each subnet
interfaces need is given as the following:

- Subnet 1: 60 interfaces.
- Subnet 2: 100 interfaces.
- Subnet 3: 15 interfaces.

For the first subnet, we need 60 interfaces that means 6 host bits. With 6 bits, we
have 26=64 addresses (Only 62 are usable). In conclusion the subnet mask will
be /26 (26 network bits and 6 host bits).

For the second subnet, we need 100 interfaces that means 7 host bits. With 7
bits, we have 27=128 addresses (Only 126 are usable). In conclusion the subnet
mask will be /25 (25 network bits and 7 host bits).

For the third subnet, we need 15 interfaces that means 5 host bits, with 5 bits we
have 25=32 addresses (Only 30 are usable). In conclusion the subnet mask will
be /27 (27 network bits and 5 host bits).

Starting with the biggest network (subnet 2). By using 164.132.62.0/25, we will
have 2 subnets (25−24=1, 21=2 subnets):
164.132.62.0/25
164.132.62.128/25

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